Question: If $f'(x)=-f(x)$ and $f(1)=2$, then $f(3)=me^n$ for some integers $m$ and $n$. What are $m$ and $n$ ? $m=~$
Explanation: If we let $y=f(x)$ and change the derivative notation, we can rewrite the differential equation so that it's separable. $\dfrac{dy}{dx}=-y$ What does it look like after we separate the variables? $\dfrac1y\,dy=-\,dx$ Let's integrate both sides of the equation. $\int\dfrac1y\,dy=\int-\,dx$ What do we get? $\ln |y|=-x+C$ What value of $C$ satisfies the initial condition $y(1)=2$ ? Let's substitute $x=1$ and $y=2$ into the equation and solve for $C$. $\begin{aligned} \ln |2| &= -1+C\\ \\ C&=1+\ln 2 \end{aligned}$ Now use this value of $C$ to express $y$ in terms of $x$. $\begin{aligned} \ln |y|&=-x+1+\ln 2\\ \\ |y|&=e^{-x+1+\ln 2}\\ \\ |y|&=e^{1-x}\cdot e^{\ln 2}\\ \\ |y|&=2e^{1-x}\\ \\ y&=\pm2e^{1-x} \end{aligned}$ Which sign satisfies the initial condition $y(1)=2$ ? We must choose the positive sign to satisfy the initial condition. $y=2e^{1-x}$ If $y(3)=me^n$, what are the integers $m$ and $n$ ? $\begin{aligned} y(3)&=2e^{1-3}\\ &=2e^{-2}\\ &=me^n \end{aligned}$ Thus, $m=2$ and $n=-2$.